spin one-half

Mixed ensemble averages

In [1], Sakurai leaves it to the reader to verify that knowledge of the three ensemble averages [S_x], [S_y],[S_z] is sufficient to reconstruct the density operator for a spin one half system.

I’ll do this in two parts, the first using a spin-up/down ensemble to see what form this has, then the general case. The general case is a bit messy algebraically. After first attempting it the hard way, I did the grunt work portion of that calculation in Mathematica, but then realized it’s not so bad to do it manually.

Attempting to do a similar set of trace expansions this way for a more general spin basis turns out to be a really bad idea and horribly messy. So much so that I resorted to \href{https://raw.githubusercontent.com/peeterjoot/mathematica/master/phy1520/spinOneHalfSymbolicManipulation.nb}{Mathematica to do this symbolic work}. However, it’s not so bad if the trace is done completely in matrix form.

So, as claimed, it’s possible to completely describe the ensemble weight factors using the ensemble averages of \( [S_x], [S_y], [S_z] \). I used the Pauli matrices instead, but the difference is just an \( \Hbar/2 \) scaling adjustment.

Pure ensemble

It turns out that doing the above is also pr. 3.10(b). Part (a) of that problem is to show how the expectation values \( \expectation{S_x}, \expectation{S_y},\expectation{S_x} \) fully determine the spin orientation for a pure ensemble.

Suppose that the system is in the state \( \ket{\BS \cdot \ncap ; + } \) as defined above, then the expectation values of \( \sigma_x, \sigma_y, \sigma_z \) with respect to this state are

We find that we don’t need to know the state of the complete system to answer questions about portions of the system, but instead just need \( \hat{\rho} \), a “probability operator” that provides all the required information about the partitioning of the system.

Pure states vs. mixed states

For pure states we can assign a state vector and talk about reduced scenarios. For mixed states we must work with reduced density matrix.

In this second example, we have more information about the left partition. That will be seen as a zero entanglement entropy in the problem set. In contrast we have less information about the first state, and will find a non-zero positive entanglement entropy in that case.

Question: Cascading Stern-Gerlach ([1] pr. 1.13)

Three Stern-Gerlach type measurements are performed, the first that prepares the state in a \( \ket{S_z ; + } \) state, the next in a \( \ket{ \BS \cdot \ncap ; + } \) state where \( \ncap = \cos\beta \zcap + \sin\beta \xcap \), and the last performing a \( S_z \) \( \Hbar/2 \) state measurement, as illustrated in fig. 1.

fig. 1. Cascaded Stern-Gerlach type measurements.

What is the intensity of the final \( s_z = -\Hbar/2 \) beam? What is the orientation for the second measuring apparatus to maximize the intensity of this beam?

so the state after filtering the \( \ket{-} \) states is \( \inv{2} \ket{-} \) with intensity (probability density) of \( 1/4 \) relative to a unit normalize input \( \ket{+} \) state to the \( \BS \cdot \ncap \) apparatus.

If \( \ket{a} = \ket{a’} \), then \( \prod_{a” \ne a’} \frac{\lr{ A – a” }}{a’ – a”} \ket{a} = \ket{a} \), whereas if it does not, then it equals one of the \( a” \) energy eigenvalues. This is a representation of the Kronecker delta function

Pick \( \ket{S_x ; \pm } = \inv{\sqrt{2}}
\begin{bmatrix}
1 \\ \pm 1
\end{bmatrix} \) as the basis for the \( S_x \) operator. Then, for the probability that the system will end up in the \( + \Hbar/2 \) state of \( S_x \), we have